# Ratio Test – Definition, Conditions, and Examples on Series

The ratio test is an important method to learn when analyzing different infinite series. It’s one of the first tests used when assessing the convergence or divergence of a given series – especially the Taylor series. The ratio test can also help us in finding the interval and radius of the interval of a power series making it a very important convergence test.

The ratio test utilizes the $\boldsymbol{n}$th and the $\boldsymbol{(n + 1)}$th term of the series. We can determine the divergence or convergence of certain series by taking the ratios of these two terms and evaluating the ratio’s limit as $\boldsymbol{n}$ approaches infinity.

This article will cover all key components we need to understand when working with series and the ratio test. In our discussion, we will:

• Establish the important conditions we need to check before we can apply the ratio test.
• Observe the different types of series that will benefit from this convergence test the most.
• Understand how the ratio test was established.
• Show you how we can apply the ratio test on different series.

Since we will be dealing with different types of series and studying their divergence as well as convergence, refresh what you know of convergent and divergent series. For now, let’s dive right into understanding the definition of the ratio test.

## What is the ratio test?

The ratio test is one of the fastest ways for us to determine whether a series is convergent or not because it only needs the $\boldsymbol{n}$th and the $\boldsymbol{(n + 1)}$th terms of the series.

In the past, we’ve learned that knowing $\lim_{n \rightarrow \infty} a_n = 0$ is not enough to conclude that the series we’re working with is convergent. This where convergence tests such as the ratio test come in handy.

Suppose that we’re working with an infinite series, $\sum_{n = 1}^{\infty} a_n$, the limit of its two successive terms’ ratio can be represented as $L$.

\begin{aligned}\lim_{n \rightarrow \infty} \dfrac{a_{ n + 1}}{a_n} &= L\end{aligned}

The ratio test utilizes the value of $\boldsymbol{L}$ to determine whether a given series is divergent or convergent. When $\boldsymbol{|L| < 1}$, the series, $\sum_{n = 1}^{\infty} a_n$, is convergent. When $\boldsymbol{|L| > 1}$, $\sum_{n = 1}^{\infty} a_n$, is divergent. Here’s the downside of using the ratio test: when $\boldsymbol{|L| = 1}$, we won’t be able to conclude anything.

## What are the ratio test rules?

Let’s rewrite what was discussed in the previous section to make the ratio test more straightforward. We’ll begin with the series, $\sum_{n =1}^{\infty} a_n$.

\begin{aligned}\sum_{n =1}^{\infty} a_n  &= a_1 + a_2 + a_3 + …+ a_n + a_{n + 1}+…\end{aligned}

We’ll use the $n$th and $(n + 1)$th terms of this series. The ratio test will depend on the value of $\boldsymbol{\lim_{n \rightarrow \infty} \left| \dfrac{a_{n+1}}{a_n}\right|}$. Let’s say we have $L$ represent this value then there are possible outcomes for $L$. Here’s a table summarizing the possible conclusions for $L$:

 RATIO TEST RULESi)                 When $\boldsymbol{L < 1}$, the series $\boldsymbol{\sum_{n =1}^{\infty} a_n}$ is absolutely convergent (and consequently, convergent).ii)               When $\boldsymbol{L > 1}$, the series $\boldsymbol{\sum_{n =1}^{\infty} a_n}$ is divergent.iii)              When $\boldsymbol{L = 1}$, the series $\boldsymbol{\sum_{n =1}^{\infty} a_n}$ may be divergent, conditionally convergent, and absolutely divergent.

The ratio test is most useful the $n$th and $(n+1)$th terms share a lot of common factors or evaluating the limit of their ratios won’t be problematic. Power series such as the Taylor series benefit from the ratio test. This is why you’ll be seeing the ratio test used in different power series in future discussions, so knowing how to apply this test correctly will come in handy later.

If you want to better understand how the ratio test was established, check the proof shown below. Otherwise, feel free to move to the next section to learn how we can use this convergence test.

### Understanding the ratio test formula proof

We’ll use the comparison test to prove the ratio test formula and we’ll break the proof into two parts: 1) when $\boldsymbol{L< 1}$ (the series converges) and 2) when $\boldsymbol{L> 1}$  (the series diverges). The goal is for us to compare, $\sum_{n = 1}^{\infty} a_n$, with a convergent geometric series.

For the first part of this proof, we’ll focus on the case when $L <r< 1$.

\begin{aligned}\lim_{n \rightarrow \infty} \left|\dfrac{a_{n + 1}}{a_n}\right| = L, \phantom{x} L < r\end{aligned}

At some point, the ratio, $\left|\dfrac{a_{n+1}}{a_n} \right|$,will be less than $r$. We can let $N$ be an integer that satisfies the inequality shown below.

\begin{aligned} \left| \dfrac{a_{n +1}}{a_n}\right| < r, \phantom{x} n \geq N\end{aligned}

This is actually how we formally define limits, so in case you need a refresher, head over to this link. For now, let’s rewrite the inequality as shown below.

\begin{aligned}|a_{n + 1}| &< |a_n|r, \phantom{x} n \geq N\end{aligned}

We’ll use this inequality to write the inequalities for $N$, $N+1$, $N +2$, and so on as shown below.

\begin{aligned}|a_{N + 1}| &< |a_N|r\\ |a_{N + 2}| &< |a_{N +1}|r < |a_N|r^2\\|a_{N + 3}| &< |a_{N +2}|r < |a_N|r^3\end{aligned}

In general, we have $|a_{N +k}| < |a_N|r^k$, where $k \geq 1$.Hence, we have the series shown below.

\begin{aligned} \sum_{k = 1}^{\infty} |a_n|r^k &= |a_N|r + |a_N|r^2 +|a_N|r^3 +|a_N|r^4 +… \end{aligned}